0
$\begingroup$

In my previous question, I asked about nontrivial sums of four powers $a^m+b^n+c^k=d^l$, and whether the nature of the solutions depend on whether $\frac{1}{m}+\frac{1}{n}+\frac{1}{k}+\frac{1}{l}$ is bigger, smaller, or equal to 1.

This heuristic (based on the Fermat Catalan conjecture, an analogous conjecture on three powers), was shown to have some rather glaring holes, with values of the reciprocal sum being arbitrarily close to 1/2 while still containing solutions. Therefore I propose a new question to repair the hole.

My new question is: Is it hard to find nontrivial integer coprime solutions to $\pm a^m\pm b^n\pm c^k=d^{lcm(m,l,k)}$ when $\frac{1}{m}+\frac{1}{n}+\frac{1}{k}+\frac{1}{lcm(m,l,k)}\le1$? Are there $m,n,k$ for which it is known there are no solutions?

Also, I would like to see parametric solutions for the cases where we do know solutions.

Improve this question
$\endgroup$
4
  • $\begingroup$ This is one case where Roman numerals work better in a title. Gerhard "Reads 'Powers Of Two' Instead" Paseman, 2020.08.13. $\endgroup$
    – Gerhard Paseman
    Commented Aug 13, 2020 at 14:33
  • $\begingroup$ Fair enough. I've changed it. The point of this question is to avoid the cases that caused an issue last time. $\endgroup$
    – Thomas
    Commented Aug 14, 2020 at 0:09
  • $\begingroup$ Setting $n=3$ in Will's example with exponents $(4,n+3,3n+3,4)$ gives a parametric solution for $(m,n,k)=(4,4,6)$ in the venue of the current question. $\endgroup$
    – Max Alekseyev
    Commented Aug 14, 2020 at 1:57
  • $\begingroup$ Let $L:=\mathrm{lcm}(m,n,)$. Then there is always a series of solutions: $$(u^n)^m-(u^m)^n + (v^{L/k})^k = v^L$$ parameterized by integer $u,v$. So, it seems that the question lacks some important restriction. $\endgroup$
    – Max Alekseyev
    Commented Aug 20, 2020 at 17:09

0

Reset to default

You must log in to answer this question.